Here is an post-mortem analysis / "how to" for the FINAL. The questions are split by the sections. At the start of each section are a few suggestions of what to look for or how to tackle the question type.

Qu1: D
A question about carbocation stability.  We have a vinyl cation (no resonance), a primary allylic cation (so has resonance stabilisation) and a simple primary carbocation. So in terms of stability ii > iii. Vinyl cations are less stable than primary carbocations so iii > i and hence overall, ii > iii > i.

Qu2: A
All about the nucleophilicity of these groups. There are two oxygen systems and a sulfur system, all are single bonded and two are negative. Let's deal with the two oxygens first.  The negative carboxylate ii is more nucleophilic than the neutral alcohol iii. Now compare the O and the S in i and iii respectively.  Since S is larger than O (one row lower in the same group of the periodic table), the sulfur is more polarisable and therefore more nucleophilic than oxygen and S-ve more nucleophilic than O-ve.  Hence in terms of nucleophilicity i > ii > iii.

Qu3: A
Leaving group ability....remember that good leaving groups need to be stable when they leave and therefore tend to be the conjugate bases of strong acids. i is the tosylate system, which is a very good leaving group, ii bromide is a good leaving group, and iii is an alcohol where HO- is a poor leaving group, so i > ii > iii.

Qu4: AB
Basicity...either think about the availability of the electrons in the base or the stability of the bases. The stronger the bases, then the more the reaction shown moves to the right. These are all based on atoms from the first row of the periodic table. In water i, there are 2 lone pairs are on an oxygen atom, in the alkoxide ii, there are 3 lone pairs on the -ve oxygen and in the amide ion iii, there are 2 lone pairs on the -ve nitrogen atom. Since N is less electronegative than O, the N system is more willing to donate its electrons and so is more basic. Therefore in terms of base strength:   iii > i > ii.

Qu5: AB
First identify the reaction.... the alcohol starting materials and reaction conditions of HBr suggest formation of an alkyl bromide via an SN1 reaction. These reactions are typically controlled by the stability of the intermediate carbocations. i would give a unfavourable primary cation, ii a secondary and and iii a secondary benzylic cation (more stable since resonance stabilised). The more stable the carbocation, the more rapidly it forms and the faster the reaction....so in terms of reactivity we have iii > ii > i.

Qu6: E
First identify the reaction.... the conditions of AgNO3 / acetone suggest an SN1 reaction (think back to the laboratory expt). A quick look at the systems shows three bromides, so we are looking at the effect of changing the alkyl group (since the leaving group is the same across the series)... since they are named the first step would be to draw the structures. i is a secondary bromide, ii is primary and iii is a benzylic bromide. These reactions are typically controlled by the stability of the intermediate carbocations, so we have iii > i > ii.

Qu7: E
Chemical shifts of the groups in question in these systems are determined by the nature of the attached groups. i is an CH3 attached to an ether type oxygen atom, so it will be near 4 ppm. ii is a CH3 group attached to a phenyl group which will cause magnetic anisotropic deshielding to about 2.2 ppm and iii is an CH in an aldehyde and is deshielded by the electronegativity and magnetic anisotropy to about 9-10 ppm. Therefore ii > iii > i.

Qu8: A
Acidity.... First let's compare i and ii (because they are very similar, both are carboxylic acids) : the difference between them is the halogen atom... the presence of the electronegative atom will stablise the conjugate base due to inductive electron withdrawal through the sigma bonds, so in terms of acidity, the more electronegative F will stabilise more than the Cl so in terms of acidity, i > ii. Now let's compare iii, it's a thiol... In terms of general principles, -SH systems are more acidic than OH systems due to the size of the S compared to the O. But that would only explain an alcohol compared to a thiol. In the carboxylic acid, there is resonance delocalisation in the conjugate base.... carboxylic acids are stronger acids that thiols. Other things that would help answer this question are knowing the pKa's (carboxylic acid are about 5, thiol about10). Therefore in terms of acidity, i > ii >iii.

Qu9: A
Acidity.... The stronger the acids, then the more the reaction shown moves to the right. i is the strong acid, sulfuric acid (pKa = -10), ii is the hydronium ion pKa = approx -2 (note that since sulfuric acid can protonate water to give hydronium we know that sulfuric acid is a stronger acid than hydronium ion) and iii is an organic acid, a carboxylic acid which are typically regarded as weak acids, pKa = 5. Therefore in terms of acidity, i > ii > iii.

Qu10: B
The number if lines = multiplicity (i.e. coupling) of the signals for each of the positions indicated and is determined by the number of neighbours that are of a different type (since H of the same type do not show coupling). i is a CH2 between two methyl groups so it has 6 neighbours and therefore appears as 7 lines. ii is a CH2 group in butane and it only has 3 neighbours of a different type (remember we don't see coupling between H of the same type) therefore we see a quartet. iii is an CH2 between two methylene (CH2) groups so it has 4 neighbours and therefore appears as 5 lines.. Therefore i > iii > ii.

Qu11: E
The reaction is the E2 elimination of an alkyl halide (draw it from the name) with different strong bases. The key here is that the larger the base, the lower the yield of the more highly substituted, more stable Zaitsev product because the larger base is hindered from abstracting the more hindered proton. So since the base sizes are iii > i > ii it means the yields of the anti-Zaitsev products is iii > i > ii.

Qu12: A
A question about alkene and alkane stability (see F2012 MT for related question in the thermodynamics section). Notice they are isomeric.  In terms of alkene stability, the general rule is the more alkyl groups on the C=C unit, the more stable it is and for chain systems, trans tend to be more stable than cis. Cyclohexane is more stable than hexene isomers because sigma bonds are stronger than pi bonds. The more stable isomer will have the less exothermic heat of combustion : i > ii >iii.

Qu 13: E
Since the SN1 goes via a carbocation, the reaction with the most stable carbocation will react fastest...but in order for the carbocation to form, there must be a good leaving group, so we are looking at the SN1 of alkyl halides. Look at the carbocations that would be formed. The most favourable carbocation is the one from II, a resonance stablised primary benzylic system.

Qu14: D
Acidity.... Both are carboxylic acids, the difference between them is the location of the halogen atom... the presence of the electronegative atom will stablise the conjugate base due to inductive electron withdrawal through the sigma bonds, so in terms of acidity, the closer the electronegative Cl is to the carboxylate group, the greater the stabilisation, so in terms of acidity, I > II.

Qu15: A
Ethanal is a simple aldehyde. The most acidic position in a simple aldehyde is the position adjacent (as in I) to the carbonyl group because this allows the -ve charge on that C to be delocalised through resonance to the more electronegative oxygen atom.

Qu16: C
Easiest to answer by pushing the curly arrows from X to derive the resonance contributor. Otherwise we use the rules for drawing resonance structures to point out that I is wrong because H atoms have moved, and III is wrong because it has a different overall charge (neutral while X is -ve).

Qu17: C
First identify the reaction.... the alcohol starting materials and reaction conditions of HBr suggest formation of an alkyl bromide via an SN1 reaction. These reactions proceed via a planar carbocation and as a result, any stereochemistry is lost and a racemic product is obtained.

Qu18: D
Boiling point is a physical property and is controlled by intermolecular forces. The key issue here would be surface area contacts... more surface area means more intermolecular forces and hence more energy required to overcome those intermolecular forces. Less branched molecules have more surface areas in contact.

Qu19: B
Need to work backwards ....looking at the product, it's an amine produced by a nucleophile substitution of an alkyl bromide that in turn was formed via a radical halogenation of a alkyl benzene. Count C atoms in the substituent. Therefore the starting material is methylbenzene.

Qu20: C
Need to work backwards ....looking at the product and the reaction conditions, it looks like we have used cyanide to introduce the nitrile group via an SN2 reaction of a tosylate. This would occur with inversion of configuration. Tosylates are made using tosyl chloride (TsCl) reacting on an alcohol (no inversion occurs during this step). Since the product is cis, we need to start from the trans-alcohol.

Qu21: A
We should work forwards .... starting from an alcohol and going to an alkene - and it's the Zaitsev product, which means we can just use a straight forward E1 elimination in a alcohol dehydration using a strong acid / heat. HBr is not the best choice as a strong acid since the nucleophilic bromide ion can resultant in competing substitution to give the alkyl bromide.

Qu22: B
We should work forwards .... halogenation of an alcohol using thionyl chloride converts an alcohol to an alkyl chloride via an SN2 reaction (i.e. inversion of stereochemistry).

Qu23: C
We should work forwards....starting from a carboxylic acid and going to an ester.... we can do this by making a nucleophilic carboxylate ion and then using an SN2 reaction using the appropriate alkyl halide. So we need a base and a benzyl halide.

Qu24: A
Need to work backwards ....looking at the product, it's an alkene - and it's probably the anti-Zaitsev product, which means we are looking at an E2 elimination which are typical of alkyl halides when heated with a base (as stated in the question). but it's the anti-Zaitsev product. In cyclic systems, the critical issue is the 180 degree arrangement of the H-C and C-LG bonds - this requires that the H and the LG in question are both axial. In this question then, the best approach is to draw the cyclohexane in the chair conformation and put the methyl group equatorial (large groups prefer to be equatorial). Then look at the position of the Br in relation to the methyl group. In order to get the required product we need trans 1,2-bromomethylcyclohexane A.

Qu25: A
We should work forwards ....starting from an alcohol and heating with a strong acid indicates an alcohol dehydration giving an alkene, an E1 elimination via a carbocation that undergoes a 1,2-methyl shift to give the more highly substituted alkene.

Qu 26: E
We should work forwards ....we are starting from an alcohol and adding 2C in the form of an alkyne. This suggests using the acetylide ion as an nucleophile, but we can't do this directly on an alcohol (due to the pKas) so we should convert the -OH into something like a halide. We have to use SN2 chemistry to do this to avoid a carbocation rearrangement. Halogenation of an alcohol using thionyl chloride converts an alcohol to an alkyl chloride via an SN2 reaction then the nucleophilic acetylide ion substitutes the -Cl.

Qu 27: D
2-methylpentane is C6H14. This rules out A, B, C (C7) and E (C4) !

Qu 28: A
The torsional angle between the methyl groups is 0 degrees for cis 1,2-dimethylcyclopropane.

Qu 29: AD
cis requires that the two methyl groups are on the same face of the ring. B, C and E are all trans-1,3-dimethylcyclohexane.

Qu30: E
The Newman projection shows a staggered conformation and shows the two indicated bonds at 60 degrees and hence gauche (this is the best, i.e. most complete term)

Qu31: C
The structures are isomeric (same MF) but what type of isomer are they? Picking up one model and comparing allows one to see that a twist around the C2-C3 bond interconverts them so they are conformational isomers.

Qu32: AC
cis-1,2-dimethylcyclohexane has a cyclohexane ring (which is regarded as being strain free, i.e. no ring strain). In the cis-1,2- isomer, one methyl will be axial and the other equatorial (i.e. they are staggered wrt each other) and the two cis methyl groups will be quite close to each other (hence Van der Waals strain). Since one of the methyl groups is axial, there will be an unfavourable interaction with the 2 C-H bonds that are also axial on that face of the ring.

Qu33: D
The molecule with the least exothermic heat of combustion is the most stable molecule. That will be cyclohexane since there is no ring strain.

Qu34: B
Monoalkyl-substituted cyclohexanes will have chair conformations where the alkyl group is either equatorial or axial. Due to unfavourable interactions when the alkyl group is axial, the equatorial form is more stable and preferred... hence they have different energies. In order for a disubstituted cyclohexane to have two chair conformations of the same energy, we would expect the same substituent such that one is axial and one is equatorial in each chair form. For 1,2-substitution, this requires that they be cis.

SPECTROSCOPY:
Use any IR information to get the functional groups. Use the H-NMR to get the number of types of H, how many of each type from the integral and what they are next to from the coupling patterns. Chemical shifts should tell you if the group is near -O- or maybe C=O groups etc.

Qu35: A
IR shows a carbonyl (i.e. C=O) at 1745 cm^-1 which is high for a typical ketone. The C-nmr peak at 171ppm suggests the C=O is a carboxylic acid or derivative. The H-nmr has 4 types of H, the triplet at 4.1 ppm, suggests a H₂C-O- group, which is next to a CH₂. The peak at 1.6ppm tells us that we have a CH₂ between a CH₂ and a CH₃. This indicates we have a -OCH₂CH₂CH₃ (i.e. an n-propyl group). Overall the pieces are -OCH₂CH₂CH₃., C=O, and -CH₃.

Qu36: BE
The H-nmr and C-nmr indicate an aromatic (H 7.14ppm and C 126, 128 and 142 ppm). This means with BD or BE. The number of types of aromatic C (3) implies it has to be BE.

Qu37: E
IR shows a carbonyl (i.e. C=O) at 1720 cm^-1 which is about normal for a typical ketone. This is supported by the C-nmr peak at 212ppm. Since there is no H-nmr peak at 9-10ppm, it is not an aldehyde. The H-nmr has 2 types of H, and the coupling and integration indicates an ethyl group.

Qu38: AC
IR shows an -OH at 3100-3500 cm^-1 and this is consistent with the exchangeable broad singlet (2.6ppm 1H) in the H-nmr. The C-nmr has 5 types of C, nothing above 70ppm means no sp2 C.

Qu39: BC
IR shows an -NH₂ at 3356 and 3184 cm^-1 and a carbonyl (i.e. C=O) at 1660 cm^-1 which is quite low for a typical ketone. The C-nmr peak at 177ppm suggests the C=O is a carboxylic acid or derivative. These facts together are suggestive of an amide. The 3 peaks in the H-nmr at 0.9 to 2.0ppm show an n-propyl group.

Qu40: AE
IR shows a very broad -OH at 2700-3300 cm^-1 and consistent with the exchangeable broad singlet (11.5ppm 1H) in the H nmr. IR also shows a carbonyl (i.e. C=O) at 1712 cm^-1. The C-nmr peak at 181ppm suggests the C=O is a carboxylic acid or derivative. These facts are suggestive of a carboxylic acid.